How fast is the train moving???

Question

The following is the equation for the Doppler shift of a sound wave of speed v reaching a moving detector.

fd = fs(v vd)/(v - vs)

In this equation, vd is the speed of the detector, and vs is the speed of the source. Also, fs is the frequency of the source, and fd is the frequency of the detector. If the detector moves toward the source, vd is positive. If the source moves toward the detector, vs is positive.

In 1845, French scientist B. Ballot first tested the Doppler shift. He had a trumpet player sound a note at 430 Hz, while riding on a flatcar pulled by a locomotive. At the same time, a stationary trumpeter played the same note. Ballot heard 3.5 beats per second. How fast was the train moving toward him?

__m/s

Answer 1

Beat frequencies work like this: you can prove that:

sin(a x) + sin(b x) = 2 cos( (a-b)x/2 ) * sin( (a+b)x/2 ).

That means that when you hear two notes of frequencies f1 and f2, what you'll hear sounds like their average frequency (f1 + f2)/2, but you'll hear it suddenly muffle whenever cos( (a-b) x / 2) = 0.

Now, cos(u) has a normal period of 2 pi. But since we register a beat whenever cos(u) is 0, we register two beats during that period (one at pi/2, one at 3 pi/2). So we hear two beats with a frequency of (f1 - f2)/2, which is a beat frequency of just f1 - f2.

So, if you hear 3.5 beats per second, that means that the difference between the two trumpet player's notes is 3.5 Hz. (A Hz is one cycle per second.)

Now, when a person travels through a static medium, and the speed of sound in the medium is fixed relative to it, the waves "bunch up" in front of her and "spread out" behind her.

Think of it this way: suppose she taps a drum with a period T -- she hits the drum and then waits T seconds or so, hits it again, waits another T seconds or so, and so on. Well, after T seconds from one hit, the sound of the drum-tap will be some distance cT ahead of her original position, and some distance -cT behind her original position. But she has moved a distance vT in that time. Which means that the wavelength ahead of her is L = cT - vT = (c - v) T, while the wavelength behind her is - cT - vT = - (c+v)T.

And if you continue that argument, with a played frequency f = 1/T  and a heard frequency f* = c / L, you'll find out that:

(f* - f) = f v / (c - v).

But f* - f is just the beat frequency you heard. Let's call it b:

b = f v / (c - v)

b (c - v) = f v

bc = bv + f v

bc / (b + f) = v.

The speed of the train, v, is the speed of sound, c, times 3.5 Hz / (433.5 Hz). The usual quoted value for the speed of sound is 343 m/s. Plug and chug.

Answer 2

1129 ft/s /430 * 3.5  =  9.19  ft/sec